<h3>
Answer:</h3>
110.98 g/mol
<h3>
Explanation:</h3>
The reaction between NaCl and Ca(OH)₂ is given by the equation;
2NaCl(aq) + Ca(OH)₂(s) → 2NaOH(aq) + CaCl₂(aq)
We are required to determine the mass of CaCl₂ produced,
We will use the following simple steps;
Step 1: Moles of NaCl and Ca(OH)₂ given
Number of moles = Mass ÷ Molar mass
Moles of NaCl
Mass of NaCl = 191 g
Molar mass NaCl = 58.44 g/mol
Number of moles = 191 g ÷ 58.44 g/mol
= 3.268 moles
= 3.27 Moles
Moles of Ca(OH)₂
Mass of Ca(OH)₂ = 74 g
Molar mass of Ca(OH)₂ = 74.093 g/mol
Number of moles = 74 g ÷ 74.093 g/mol
= 0.998 mole
= 1.0 mole
However, from the equation 2 moles of NaCl requires 1 mole of Ca(OH)₂
Therefore, from the amount of reactants available NaCl was in excess and Ca(OH)₂ is the limiting reactant .
Step 2: Moles of CaCl₂ produced
From the equation
1 mole of Ca(OH)₂ reacts with NaCl to produce 1 mole of CaCl₂
Therefore; the mole ratio of Ca(OH)₂ to CaCl₂ is 1: 1
Thus;
Moles of CaCl₂ produced is 1.0 moles
Step 3: Mass of CaCl₂ produced
Moles of CaCl₂ = 1.0 mole
Molar mass CaCl₂ = 110.98 g/mol
But; mass = number of moles × Molar mass
Therefore;
Mass of CaCl₂ = 1.0 mole × 110.98 g/mol
= 110.98 g CaCl₂
